1) $\left \{ {{4x+4y= x^2+y^2 +2y+ 4} \atop {x( y+1) = x - 3}} \right.$ ⇔$\left \{ {{4(x+y)=(x+y)^2+4} \atop {xy=-3}} \right.$
⇔$\left \{ {{(x+y-2)^2=0} \atop {xy=-3}} \right.$ ⇔$\left \{ {{x+y=2} \atop {xy=-3}} \right.$
⇔$\left \{ {{y=2-x} \atop {x(2-x)=-3}} \right.$ ⇔\(\left[ \begin{array}{l}x=3;y=-1\\x=-1;y=3\end{array} \right.\)
2)$\left \{ {{x^2+ y^2+ x+y = -2} \atop { x^2- y^2 + 2x- 4y = 3}} \right.$ ⇔$\left \{ {{x^2+ y^2+ x+y = -2} \atop {(x+1)^2-(y+2)^2=0}} \right.$
⇔$\left \{ {{ x^2+ y^2+ x+y = -2} \atop {(x-y-1)(x+y+3)=0}} \right.$ ⇔$\left \{ {{x^2+ y^2+ x+y = -2} \atop {\left[ \begin{array}{l}y=x-1\\y=-x-3\end{array} \right.}} \right.$
TH1: $\left \{ {{x^2+ y^2+ x+y = -2} \atop {y=x-1}} \right.$ ⇔$\left \{ {{x^2+(x-1)^2+x+x-1=-2} \atop {y=x-1}} \right.$ ⇔$\left \{ {{x^2=-2} \atop {y=x-1}} \right.$ (vô nghiệm)
TH2: $\left \{ {{x^2+ y^2+ x+y = -2} \atop {y=-x-3}} \right.$ ⇔$\left \{ {{x^2+(x+3)^2+x+(-x-3)=-2} \atop {y=-x-3}} \right.$ ⇔$\left \{ {{2x^2+6x+8=0} \atop {y=-x-3}} \right.$ (vô nghiệm)
3)Đk: y≥0
Đặt $\sqrt{y}$ =t (t$\geq$ 0)
Ta có: $\left \{ {{x^2+xt=2} \atop {4t^2+3xt=-2}} \right.$ ⇔$\left \{ {{x^2+xt=2} \atop {4t^2+3xt=-(x^2+xt)}} \right.$
⇔$\left \{ {{x^2+xt=2} \atop {x^2+4xt+4t^2=0}} \right.$ ⇔$\left \{ {{x^2+xt=2} \atop {(x+2t)^2=0}} \right.$
⇔$\left \{ {{4t^2-2t^2=2} \atop {x=-2t}} \right.$ ⇔$\left \{ {{t=±1} \atop {x=-2t}} \right.$
Do t$\geq$ 0 nên t=1⇒x=-2;y=1
